Muhammad Haris Rao
Proposition: If $n \in \mathbb{Z}_{> 0}$ then $\Gamma(n) = (n - 1)!$.
Proof. If $n = 1$ then we have $$ \Gamma(1) = \int_0^\infty e^{-t} \, dt = \left[ - e^{-t} \right]_0^\infty = 1 = 0! $$ which is what we want. Now we induct on this case. If $\Gamma(n) = (n - 1)!$ for some $n \in \mathbb{Z}_{> 0}$, then by the recurrence relation $$ \Gamma(n + 1) = n \Gamma(n) = n (n - 1)! = n! $$ which completes the induction step. So we can conclude that $\Gamma(n) = (n - 1)!$ for all $n \in \mathbb{Z}_{> 0}$. $\blacksquare$
Proposition: If $n \in \mathbb{Z}_{\ge 0}$ then \begin{align*} \Gamma \left( n + \frac{1}{2} \right) = \frac{(2n)!}{4^n n!} \sqrt{\pi} \\ \Gamma\left( -n + \frac{1}{2} \right) = \frac{(-4)^n n!}{(2n)!} \sqrt{\pi} \end{align*}
Proof. First we take care of the $n = 0$ case. That is, we want to compute $\Gamma(1/2)$. By reflection formula, $$ \Gamma\left( \frac{1}{2} \right)^2 = \Gamma\left( \frac{1}{2} \right) \Gamma\left( 1 - \frac{1}{2} \right) = \frac{\pi}{\sin{ (\pi/2) }} = \pi $$ so $\Gamma(1/2) = \pm \pi$. But see that $$ \Gamma(1/2) = \int_0^\infty \frac{e^{-t}}{\sqrt{t}} \, dt \ge 0 $$ since the integrand is non-negative. So it must be $\Gamma(1/2) = \sqrt{\pi}$, which is what we want.
Now to take care of the case where $n \in \mathbb{Z}_{\ge 0}$. The $n = 0$ case has been demonstrated already, so we have to induct on this case. Indeed, if $n \in \mathbb{Z}_{\ge 0}$ is such that $$ \Gamma \left( n + \frac{1}{2} \right) = \frac{(2n)!}{4^n n!} \sqrt{\pi} $$ Then the $n + 1$ case will be $$ \Gamma\left( (n + 1) + \frac{1}{2} \right) = \left( n + \frac{1}{2} \right) \Gamma \left( n + \frac{1}{2} \right) = \frac{2n(2n + 1)}{4 n} \left( \frac{(2n)!}{4^n n!} \sqrt{\pi} \right) = \frac{(2(n + 1))!}{4^{n + 1} (n + 1)!} \sqrt{\pi} $$ which completes the induction step. So the desired result holds for all $n \in \mathbb{Z}_{\ge 0}$.
Now for the case where $n \in \mathbb{Z}_{\le 0}$. Equivalently, we will evaluate $\Gamma( - n + 1/2 )$ for $n \in \mathbb{Z}_{\ge 0}$. Suppose that we know that $$ \Gamma\left( -n + \frac{1}{2} \right) = \frac{(-4)^n n!}{(2n)!} \sqrt{\pi} $$ for some $n \in \mathbb{Z}_{\ge 0}$. Then it follows from the recurrence relation that $$ \Gamma \left( -(n + 1) + \frac{1}{2} \right) = \frac{\Gamma\left( -n + \frac{1}{2} \right)}{-(n + 1) + \frac{1}{2}} = \left( \frac{-2 }{2n + 1} \right) \Gamma\left( -n + \frac{1}{2} \right) = \left( \frac{-4 n}{2n(2n + 1)} \right) \frac{(-4)^n n!}{(2n)!} \sqrt{\pi} = \frac{(-4)^{n + 1} (n + 1)!}{\left( 2(n + 1) \right)!} \sqrt{\pi} $$ By principle of mathematical induction, this proves the second claim. $\blacksquare$
Proposition: If $n \in \mathbb{Z}_{> 0}$, then $$ \Gamma^\prime(n) = (n - 1)! \left( - \gamma + H_{n - 1} \right) $$ where $H_n = 1 + 1/2 + 1/3 + \cdots + 1/n$ is the $n$th harmonic number, and $\gamma$ is the Euler-Mascheroni constant.
Proof. Recall the digamma function $\psi$, defined as the logarithmic derivative of the Gamma function and having representation $$ \frac{\Gamma^\prime(s)}{\Gamma(s)} = \psi(s) = - \frac{1}{s} - \gamma + \sum_{k = 1}^\infty \frac{s}{k(s + k)} $$ valid for all $s \in \mathbb{C} - \mathbb{Z}_{\le 0}$. We can find a recurrence relation for this. See that if $s \in \mathbb{C} - \mathbb{Z}_{\le 0}$, then \begin{align*} \Gamma^\prime \left( s + 1 \right) &= \lim_{h \to 0} \frac{\Gamma(s + 1 + h) - \Gamma(s + 1)}{h} \\ &= \lim_{h \to 0} \frac{(s + h) \Gamma(n + h) - s \Gamma(s)}{h} \\ &= s \lim_{h \to 0} \frac{\Gamma(s + h) - \Gamma(s)}{h} + \lim_{h \to 0} \Gamma(s + h) \\ &= s \Gamma^\prime(s) + \Gamma(s) \\ \end{align*} Dividing through by $\Gamma(s + 1)$ gives \begin{align*} \psi(s + 1) = \frac{\Gamma^\prime(s + 1)}{\Gamma(s + 1)} = \frac{s \Gamma^\prime(s) + \Gamma(s)}{s \Gamma(s)} = \psi(s) + \frac{1}{s} \end{align*} From this, we obtain the recurrence relation \begin{align*} \psi(n) = \psi(1) + H_{n - 1} \end{align*} when $n \in \mathbb{Z}_{> 0}$. The value of $\psi(1)$ can be computed explicitly as $$ \psi(1) = - 1 - \gamma + \sum_{k = 1}^\infty \frac{1}{k(k + 1)} = -1 - \gamma + \sum_{k = 1}^\infty \left( \frac{1}{k} - \frac{1}{k - 1} \right) = -1 - \gamma + 1 = - \gamma $$ where the sum evaluates to 1 by considering it as a telescoping series. We have now $\psi(n) = -\gamma + H_{n - 1}$, and consequently, $$ \Gamma^\prime (n) = \Gamma(n) \psi(n) = (n - 1)! \left( - \gamma + H_{n - 1} \right) $$ which is what we wanted. $\blacksquare$
Proposition: If $n \in \mathbb{Z}_{\ge 0}$, then \begin{align*} \Gamma^\prime\left( n + \frac{1}{2} \right) &= \sqrt{\pi} \left( \frac{(2n)!}{4^n n!} \right) \left( 2 H_{2n} - H_{n} - \gamma - \log{4} \right) \\ \Gamma^\prime\left( -n + \frac{1}{2} \right) &= \sqrt{\pi} \left( \frac{(-4)^n n!}{(2n)!} \right) \left( 2 H_{2n} - H_n - \gamma - \log{4} \right) \end{align*}
Proof. First to evaluate $\Gamma^\prime(1/2)$. Using the fact that $$ \gamma = \lim_{N \to \infty} \left( \sum_{n = 1}^N \frac{1}{n} - \log{N} \right) $$ we can evaluate $\psi(1/2)$ as \begin{align*} \psi \left( \frac{1}{2} \right) &= - \frac{1}{1/2} - \gamma + \sum_{k = 1}^\infty \frac{1/2}{n(n + 1/2)} \\ &= -2 - \gamma + 2 \sum_{n = 1}^\infty \frac{1}{2n(2n + 1)} \\ &= -2 - \gamma + 2 \lim_{N \to \infty} \left( \sum_{n = 1}^N \frac{1}{2n} - \sum_{n = 1}^N \frac{1}{2n + 1} \right) \\ &= -2 - \gamma + 2 \lim_{N \to \infty} \left( \frac{1}{2} \sum_{n = 1}^N \frac{1}{n} - \sum_{n = 1}^{2N} \frac{1}{n} - \frac{1}{2N + 1} + 1 + \sum_{n = 1}^N \frac{1}{2n} \right) \\ &= -2 - \gamma + 2 \lim_{N \to \infty} \left( \sum_{n = 1}^N \frac{1}{n} - \log{N} - \sum_{n = 1}^{2N} \frac{1}{n} + \log{2N} - \log{2} - \frac{1}{2N + 1} + 1 \right) \\ &= -2 - \gamma + 2 \lim_{N \to \infty} \left( \sum_{n = 1}^N \frac{1}{n} - \log{N} \right) - 2 \lim_{N \to \infty} \left( \sum_{n = 1}^{2N} \frac{1}{n} - \log{2} - \log{2N} \right) - 2\lim_{N \to \infty} \frac{1}{2N + 1} + 2 \\ &= -2 - \gamma + 2 \gamma - 2 \gamma - \log{4} - 0 + 2 \\ &= - \gamma - \log{4} \end{align*} Then using the values of $\Gamma$ at half integer values, $$ \Gamma^\prime \left( \frac{1}{2} \right) = \Gamma \left( \frac{1}{2} \right) \psi \left( \frac{1}{2} \right) = - \sqrt{\pi} \left( \gamma + \log{4} \right) $$ Recall the relation $\psi(s + 1) = \psi(s) + 1/s$ for $s \in \mathbb{C} - \mathbb{Z}_{\le 0}$. Then we have for $n \in \mathbb{Z}$ $$ \psi\left( n + \frac{1}{2} \right) = \psi \left( (n - 1) + \frac{1}{2} \right) + \frac{1}{n - 1/2} = \psi \left( \left( n - 1 \right) + \frac{1}{2} \right) + \frac{2}{2n - 1} $$ A simple application of mathematical induction will prove that if $n \in \mathbb{Z}_{\ge 0}$, then $$ \psi\left( n + \frac{1}{2} \right) = \psi \left( \frac{1}{2} \right) + 2 \sum_{k = 1}^n \frac{1}{2k - 1} = - \gamma - \log{4} + 2 \sum_{k = 1}^n \frac{1}{2k - 1} = - \gamma - \log{4} + 2 \left( H_{2n} - \frac{1}{2} H_n \right) $$ Hence, the derivative of $\Gamma$ at $n \in \mathbb{Z}_{\ge 0}$ is \begin{align*} \Gamma^\prime \left( n + \frac{1}{2} \right) &= \Gamma \left( n + \frac{1}{2} \right) \psi\left( n + \frac{1}{2} \right) = \sqrt{\pi} \left( \frac{(2n)!}{4^n n!} \right) \left( 2 H_{2n} - H_n - \gamma - \log{4} \right) \\ \end{align*} This proves the first claim.
Now to evaluate $\psi(-n + \frac{1}{2})$ when $n \in \mathbb{Z}_{\ge 0}$. Since $\psi(s + 1) = \psi(s) + 1/s$, it follows that $\psi(s) = \psi(s + 1) - 1/s$ for all $s \in \mathbb{C} - \mathbb{Z}_{\le 0}$. Hence, we have for $n \in \mathbb{Z}_{> 0}$ $$ \psi \left( -n + \frac{1}{2} \right) = \psi \left( -(n - 1) + \frac{1}{2} \right) - \frac{1}{-n + 1/2} = \psi \left( -(n - 1) + \frac{1}{2} \right) + \frac{2}{2n - 1} $$ It is easily proven then from this reecurrence that $$ \psi \left( -n + \frac{1}{2} \right) = \psi \left( \frac{1}{2} \right) + 2 \sum_{k = 1}^n \frac{1}{2k - 1} $$ The expression on the right is exactly what we had for the value of $\psi \left( n + 1/2 \right)$. Hence, we have for all $n \in \mathbb{Z}_{\ge 0}$ that $\psi(-n + 1/2) = \psi(n + 1/2)$. So we can say then that \begin{align*} \Gamma \left( - n + \frac{1}{2} \right) &= \Gamma \left( -n + \frac{1}{2} \right) \psi \left( -n + \frac{1}{2} \right) \\ &= \frac{(-4)^n n!}{(2n)!} \sqrt{\pi} \psi \left( n + \frac{1}{2} \right) \\ &= \sqrt{\pi} \left( \frac{(-4)^n n!}{(2n)!} \right) \left( 2 H_{2n} - H_n - \gamma - \log{4} \right) \\ \end{align*} which is the second claim. $\blacksquare$
Recall that if $s \in \mathbb{C} - \mathbb{Z}_{\le 0}$, then $$ \Gamma^\prime(s + 1) = s \Gamma^\prime(s) + \Gamma(s) $$ We would like to generalise such a recurrence for higher derivatives.
Theorem: If $n \in \mathbb{Z}_{> 0}$, then
Proof. If $s \in \mathbb{C} - \mathbb{Z}_{\le 0}$, then \begin{align*} \Gamma^{\prime\prime} (s + 1) &= \lim_{h \to 0} \frac{\Gamma^\prime(s + 1 + h) - \Gamma(s + 1)}{h} \\ &= \lim_{h \to 0} \frac{(s + h) \Gamma^\prime(s + h) + \Gamma(s + h) - s \Gamma^\prime(s) - \Gamma(s)}{h} \\ &= s \lim_{h \to 0} \frac{\Gamma^\prime (s + h) - \Gamma^\prime(s)}{h} \end{align*}